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Show that $$\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \frac{a\sqrt{3}+b-\pi}{6},$$ where $a$ and $b$ are constants to be found.

Answer is: $$\frac{24\sqrt3-48-\pi}{6}$$

Thank you in advance!

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You need to give the limits for the integral, so you obtain a constant. Otherwise, the integral of the expression is just the new expression you have. Also, LaTeX would be better for clearer representation of the integrals. – NasuSama May 7 '13 at 14:59
int <?> {4x-5} over {sqrt{3+2x-x^2{}} } dx – Hahanotfunny May 7 '13 at 15:06
Oops, I dont know how to go about using LaTeX (I have Wiki-ed it). Im using OpenOffice Formula. I hope its better. – Hahanotfunny May 7 '13 at 15:08
Thanks @Inceptio for making it clearer to read – Hahanotfunny May 7 '13 at 15:15
Check the edit history. It wasn't me. – Inceptio May 7 '13 at 15:16

$$I=\int_0^1\frac{4x-5}{\sqrt{3+2x-x^2}}dx = \int_0^1\frac{4x-5}{\sqrt{4-(x-1)^2}}dx$$

Let $x-1=2\sin\theta\implies dx=2\cos\theta d\theta$

If $x=0, \sin\theta=\frac12, \theta=\frac\pi6$

If $x=1, \sin\theta=0, \theta=0$

$$\text{So,}I=\int_{\frac\pi6}^0\frac{4(2\sin\theta+1)-5}{2\cos\theta}\cdot2\cos\theta d\theta$$ $$=8\int_{\frac\pi6}^0\sin\theta d\theta-\int_{\frac\pi6}^0\theta d\theta$$ $$=8[-\cos\theta]_{\frac\pi6}^0-\left(0-\frac\pi6\right)$$

Can you take it home from here?

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(+1) My favourite. – Inceptio May 7 '13 at 15:57
I managed to solve it using substitution method. Your method is interesting, I have not come across it before. Thanks for the suggestion! – Hahanotfunny May 7 '13 at 16:06
@Hahanotfunny, my pleasure.Trigonometric identities are often panacea to me for integration. – lab bhattacharjee May 7 '13 at 16:09
For $(x-a)^2+b^2,$ put $x-a=b\tan\theta$ or $b\cot\theta$ For $(x-a)^2-b^2,$ put $x-a=b\sec\theta$ or $b\csc\theta$ For $b^2-(x-a)^2,$ put $x-a=b\sin\theta$ or $b\cos\theta$ – lab bhattacharjee May 7 '13 at 16:17
@Hahanotfunny, we have used the Principal values (…) of $\arcsin$ for which $\sqrt{4-(x-1)^2}=+\cos \theta$. As both the values of $\sin\theta\ge 0$ we could take the angles in the second Quadrant namely, $\arcsin 0=\pi$ and $\arcsin \frac12=\frac{5\pi}6,$ then $\sqrt{4-(x-1)^2}$ would become $-\cos\theta$ as $\cos\theta<0$ in the second Quadrant – lab bhattacharjee May 8 '13 at 15:33

Note first that the derivative of the quadratic in the denominator is $-2x+2$. It would be great if the numerator were $4x-4$, because then we could set $u=3+2x-x^2$, $du=(-2x+2)dx$, and write the indefinite integral as

$$\int\frac{-2}{\sqrt u}du=-2\int u^{-1/2}du\;.$$

Unfortunately, the numerator is actually $4x-5=-2(-2x+2)-1$. The trick is to split the integral in two:


We’ve already sorted out how to deal with the first of these integrals, so that leaves only the second. Complete the square:




which can be handled with a standard trig substitution.

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3+2x-x^2=-(x^2-2x-3)=-\big((x-1)^2-4\big)=2^2-(x-1)^2\;, Gonna highlight this for future reference. Haha. I kept on thinking that its not possible to have a negative in a square root unless its Complex. Thanks! – Hahanotfunny May 7 '13 at 16:11
@Hahanotfunny: You’re welcome! – Brian M. Scott May 7 '13 at 16:14

On solving we will find that it is equal to -$$-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$

Now if you put the appropriate limits I guess you'll get your answer.

First of all write $$4x-5 = \mu \frac{d(3+2x-x^{2})}{dx}+\tau(3+2x-x^{2})$$

We will find that $\mu=-2$ and $\tau=-1$.

$$\int\frac{4x-5}{\sqrt{3+2x-x^{2}}}=\mu\int\frac{d(3+2x-x^{2})}{\sqrt{3+2x-x^{2}}}+\tau\int\frac{dx}{\sqrt{3+2x-x^{2}}}$$$$= -2(2\sqrt{(3+2x-x^{2}})+-1\left(\int\frac{d(x-1)}{\sqrt{2^{2}-(x-1)^{2}}}\right) $$$$=-4\sqrt{3+2x-x^{2}}-\sin^{-1}(\frac{x-1}{2})$$

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The limit is 0 to 1. Sorry about that. – Hahanotfunny May 7 '13 at 15:14
How did you make the square root to become positive? I couldnt carry on at -2 int (2-2x) /<sqrt{3+2x-x^2{}} > - arcsin((x-1)/2) – Hahanotfunny May 7 '13 at 15:31

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